The equation of a circle $C$ is $x^2+y^2+4x-14y+4 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+4x) + (y^2-14y) = -4$ $(x^2+4x+4) + (y^2-14y+49) = -4 + 4 + 49$ $(x+2)^{2} + (y-7)^{2} = 49 = 7^2$ Thus, $(h, k) = (-2, 7)$ and $r = 7$.